Pneumatica ROSS

Bulletin 371C

PneumaticReference

Data

CONTENTS

PNEUMATIC PRINCIPLES .................................. 2–4

GLOSSARY OF USEFUL TERMS ........................... 5

VALVE RESPONSE TIME ........................................ 6

FLOW COEFFICIENTS ............................................ 7

STANDARDIZED TESTING ..................................... 8

FLOW CHART FOR Cv = 1 ...................................... 9

CYLINDER VOLUME.............................................. 10

VALVE SIZING ....................................................... 11

PNEUMATIC SYMBOLS .................................. 12–14

PORT IDENTIFICATION................................... 14–15

CONVERSION FACTORS ...................................... 15

ROSS CONTROLS® 2

Pneumatic Principles

Pneumatics is the study of the mechanical properties ofgases. In industrial applications the gas involved is mostcommonly compressed air. The mechanical propertieswith which we are most concerned are pressure, volumeand temperature. The relationships between these prop-erties, and the meanings of some basic pneumaticterminology are discussed below.

Force and Pressure. Consider the diagram of thecylinders below.

Figure 1

When the force of 60 pounds is applied to the piston, theair in the cylinder is compressed. This compressioncontinues until the upward force on the piston is thesame as the downward force, that is, until the forces arein equilibrium.

The force on each square inch of the piston is:

force on piston = 60 lb = 15 lb per in2 (psi)area of piston 4 in2

This force per unit area is called pressure. The termsforce and pressure must not be confused. Force isexpressed in units such as pounds, tons, kilograms, etc.,while pressure is expressed as pounds per square inch,tons per square mile, kilograms per square meter, etc.

In general terms, then:

pressure = force area

Conversely, the force exerted on or by an area is:

force = pressure x area

Pascal’s Law. Pressure applied to a confined fluid istransmitted undiminished in all directions. In the discus-sion of Figure 1 we learned that the force applied to thepiston produces a pressure of 15 psi (pounds per squareinch). Pascal’s Law simply says that a pressure of 15 psiis applied to all surfaces containing the air, i.e., the wallsand bottom of the cylinder as well as the piston surfaceitself.

60 lb

Piston area4 sq in

60-lb upward forceresulting fromcompression

Atmospheric Pressure. The pressure produced on theearth's surface by the weight of the air surrounding theearth was measured by the 17th-century scientist,Evangelista Torricelli. He filled a 36-inch glass tube(sealed at one end) with mer-cury, and placed his finger overthe open end of the tube. Hethen stood the tube on end in adish of mercury and removedhis finger. Some mercuryflowed from the tube into thedish, but a column of mercury30 inches tall remained in thetube. The pressure of the at-mosphere, he reasoned, is suf-ficient to support a column ofmercury 30 inches high. Morerefined tests have shown thisatmospheric pressure to beequal to 14.69 pounds persquare inch.

30"Hg

Vacuum

Figure 2

Gauge Pressure. Atmospheric pressure is a part of oureveryday environment. We don’t notice the pressure onus, and it is convenient to think of atmospheric pressureas “zero” pressure. So we measure pressure starting atthis zero point, and a pressure so measured is calledgauge pressure. A common abbreviation for gaugepressure measurements is psig (pounds per square inchgauge).

Absolute Pressure. When dealing with the effects ofpressure on gases it is necessary to consider the totalpressure on the gas. We can no longer ignore the effectof atmospheric pressure as we do when we measuregauge pressure. This total pressure acting on a gas iscalled absolute pressure.

absolute pressure = gauge pressure + 14.69

You will recognize 14.69 as the pounds-per-square-inchmeasure of atmospheric pressure.

To distinguish absolute pressure from gauge pressuremeasurements, we use units such as psia (pounds persquare inch absolute).

Absolute Temperature. Absolute temperature is basedon a theoretical “lowest” temperature. This lowest tem-perature is the zero point for the two absolute tempera-ture scales in common use, the Rankine scale and theKelvin scale.

The Rankine scale is based on the Fahrenheit degree.

Rankine temperature = Fahrenheit temperature + 459.67

Thus, 0°F = 459.67°R, although for most practical ap-

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plications it is taken as 460°R. The Kelvin scale ofabsolute temperature is based on the Celsius degree.

Kelvin temperature= Celsius temperature + 273.16

Thus, 0°C = 273.16°K. Again, for most practical pur-poses it is taken as 273°K.

Perfect Gas Laws. The relationships between pres-sure, volume, and temperature are expressed in threebasic “laws” which we will discuss below. These lawsapply to an ideal gas, but are also remarkably accuratefor the mixture of gases making up our air.

In applying these laws it is important to remember thatonly absolute values of pressure and temperature maybe used.

Boyle's Law. This law expresses the relationshipbetween pressure and volume when temperature is heldconstant. According to this law, the volume of the gas ina container is inversely proportional to the absolutepressure on the gas, i.e.,

P1V1 = P2V2

Let us apply this law to find the volume of the gas in thecylinder below after compression.

Figure 3

The pressure acting on the cylinder at the left is simplyatmospheric pressure. The absolute pressure on thecylinder after it is compressed is:

P2 = (60 ÷ 4) + 14.7 = 29.7 psia

With this information we can use Boyle's Law to find thecompressed volume.

14.7 x 120 = 29.7 x V2

V2 = 14.7 x 120 ÷ 29.7

= 59.4 cubic inches

When a gas is compressed it gets warmer, so the abovecalculation would be correct only after the compressed

60 lbPiston area

4 sq in

V1

120 cu in V2

P1 = 14.7 psia

P2

gas has been allowed to cool to the temperature it hadbefore compression.

Charles’ Law. This law expresses the relationshipbetween volume and temperature when pressure is heldconstant. According to this law, the volume of the gas inan expandable container is directly proportional to theabsolute temperature of the gas, i.e.,

V1T2 = V2T1

Let us take the cylinder in Figure 3 and specify that thetemperature of the gas is 60°F. If we apply heat to thecontainer, as depicted in Figure 4 below, so that thetemperature of the gas is raised to 100°F, what will be thenew volume of the gas. Note that the load on the pistondoes not change, and therefore the pressure on the gasis constant as required by Charles’ Law.

Figure 4

The absolute temperatures are:

T1 = 60°F + 460F° = 520°R

T2 = 100°F + 460F° = 560°R

The volume of the compressed gas as we determined itfrom Figure 3 is V1 = 59.4 cubic inches. When we applyCharles’ Law to this information we have:

59.4 x 560 = V2 x 520

V2 = 59.4 x 560 ÷ 520

= 64.0 cubic inches

Gay-Lussac’s Law. The third gas law states that if thevolume of a gas is held constant (i.e., confined in a rigidcontainer) the absolute pressure of the gas is directlyproportional to its absolute temperature. Thus,

P1T2 = P2T1

A closed container when heated builds up considerableinternal pressure, and can lead to an explosion of thecontainer. This is Gay-Lussac’s Law at work!

60 lb

V1 T1

60 lb

V2 T2

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General Gas Law. In practical applications all three ofthe variables — pressure, volume, temperature — maychange. The three gas laws we have discussed can beconsolidated into a single general law which provides forchanges in all three variables. This general law is simply:

P1V1T2 = P2V2T1

Let us apply this general law to the following problem.

Figure 5

Initially, the force on the cylinder’s piston is 400 lb. Thiscompresses the gas in the cylinder to a volume of 120cubic inches at a temperature of 60°F. If the gas in thecylinder is heated to 200°F, what force must be on thepiston to compress the gas further to a volume of 100cubic inches?

Because the piston has an area of 4 square inches, theabsolute pressure on the gas initially is:

P1 = (400 ÷ 4) + 14.7 = 114.7 psia,

and the absolute temperatures are:

T1 = 60 + 460 = 520°R

T2 = 200 + 460 = 660°R.

In order to determine the force that must be on the pistonin order to compress the heated gas to 100 cubic inches,we must first find the value of P2. To do this we apply thegeneral gas law, P1V1T2 = P2V2T1.

114.7 x 120 x 660 = P2 x 100 x 520

P2 = (114.7 x 120 x 660) ÷ (100 x 520)

= 174.7 psia

P2 in terms of the force F acting on the piston is:

P2 = (F ÷ 4) + 14.7

Then we have,

174.7 = (F ÷ 4) + 14.7

F ÷ 4 = 174.7 - 14.7 = 160 psig

F = 160 x 4 = 640 lb.

Standard Air and Free Air. In the United States thevolume of air is most commonly measured by the cubicfoot. Of course the actual amount of air in a cubic footdepends on factors such as pressure and temperature,so it has become necessary to define a standard cubicfoot (scf) of air. For most industrial applications this is acubic foot of air at a pressure of one atmosphere (14.69psia), a temperature of 68°F, and a relative humidity of36 per cent. The weight of a standard cubic foot of air is0.0762 lb; 1 lb of standard air occupies 13.1 cubic feet.

In actual practice one seldom encounters standard air,but is dealing with “free” air, that is, the air that weencounter in everyday experience. Fortunately, unlesswe are on a mountaintop or working near a blast furnace,free air is close enough to standard air so that we do nothave to make corrections for the small differences.

Compressed Air. The flow of air through a valve or anypart of a pneumatic system is usually measured instandard cubic feet per minute (scfm). Often it is neces-sary to know what this volume of air is when com-pressed. Boyle’s law solves the problem easily.

For example, if the air flow through a valve is 50 scfm,what is the actual volume of the air if the pressure is 100psig? Remembering that Boyle’s law requires that weuse absolute pressures, we have:

14.7 x 50 = (100 + 14.7) x V

where V is the volume of the compressed air. We obtain,

V = (14.7 x 50) ÷ (100 + 14.7)

V = 6.4 cubic feet per minute.

Conversely, we could ask for the pressure required to fillan automobile tire with 2 cubic feet of free air if the tirehas a volume of 0.5 cubic foot. Again, from Boyle’s lawwe obtain:

14.7 x 2 = (P + 14.7) x 0.5

where P is the gauge pressure required to fill the tire.

P + 14.7 = (14.7 x 2) ÷ 0.5

P + 14.7 = 58.8

P = 44.1 psig.

400 lb

? lbPiston area4 sq in

120 cu in

60 F100 cu in

200 F

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Absolute Pressure: The sum of atmospheric pressureand gauge pressure. Designated as psia (pounds persquare inch absolute).

Air Receiver: A container in which air is stored underpressure as a source of pneumatic power.

Ambient Temperature: The temperature of the imme-diate environment.

Atmospheric Pressure: The pressure exerted by theatmosphere. At sea level this pressure is 14.69 poundsper square inch absolute.

Bar: A unit of pressure measurement equal to 105

newtons per square meter or 14.50 pounds per squareinch.

Celsius, Degree: A unit of temperature measurementabbreviated °C. Celsius temperatures are calculatedfrom Fahrenheit temperatures by the following formula:

C = 5(F 9 32)

Closed Center: Pertains to three-position valves; allports are closed when the valve is in the center position.

Detent: A device for retaining movable parts in one ormore fixed positions; usually a spring-loaded devicefitting into a depression. Positions of parts are changedby exerting sufficient force to overcome the detentspring, or by releasing the detent.

Directional Control Valve: A valve whose function is todirect or prevent flow through selected passages.

Flow Rate: The volume or weight of a fluid passingthrough a conductor per unit of time. For pneumaticsystems, flow rate is most often expressed as standardcubic feet per minute (scfm).

Fluid: A liquid or a gas.

Fahrenheit, Degree: A unit of temperature measure-ment abbreviated °F. Fahrenheit temperatures are cal-culated from Celsius temperatures by the followingformula:

F = 1.8C + 32

Four-Way Valve: Now designated as a 4/2 valve; afour-port, two position valve.

Free Air: Air at atmospheric pressure.

Gauge Pressure: Pressure above or below atmo-spheric pressure. Designated as psig (pounds per squareinch gauge).

Holding Power: Amount of electric power required tomaintain solenoid in its actuated position.

Impulse Signal: A briefly applied pneumatic or electricsignal; a momentary signal.

Inrush Power: Amount of electric power required dur-ing time solenoid plunger is moving.

Media: The fluids used in a fluid power system. In apneumatic system they are gases such as air, nitrogen,or various inert gases.

Media Temperature: The temperature of the fluid withina valve or other device.

Micron: A measurement equal to one-millionth of ameter or about 0.00004 inch.

Momentary Signal: A briefly applied pneumatic orelectric signal; an impulse signal.

Normally Closed: A term used to describe a valvewhich blocks the flow of supply air when the valve is notactuated.

Normally Open: A term used to describe a valve whichallows supply air to flow only when the valve is notactuated.

Open Center: Pertains to three-position valves; outletports are connected to exhaust when valve is in thecenter position.

Pressure: A measure of force per unit area; oftenexpressed as pounds per square inch (psi) or bar.

Pressure Range: The range of inlet pressures withwhich a device can operate satisfactorily.

Signal: A fluid or electric command to the valve actuatorcausing the valve to change position.

Silencer: A device used to reduce the sound levelproduced by an exhausting gas.

Standard Air: Air at a temperature of 68°F, a pressureof 14.69 pounds per square inch absolute (psia), and arelative humidity of 36 per cent (0.0750 pounds percubic foot). In gas industries the temperature of stan-dard air is usually specified as 60°F.

Three-Way Valve: Now designated as a 3/2 valve; athree-port, two-position valve.

Two-Way Valve: Now designated as a 2/2 valve; a two-port, two-position valve. Also know as a straightwayvalve.

Vacuum: Pressure less than atmospheric pressure.

Glossary of Useful Terms

ROSS CONTROLS® 6

Valve Response Time

Most pneumatic applications call for a valve to be usedto control the repeated filling and exhausting of a device(cylinder, clutch, etc.) having a certain volume. The timerequired to fill or exhaust this volume is called the valveresponse time. The time to fill a volume is usuallydifferent from the time to exhaust the volume. Bothtimes can be found by using the following formula:

Valve Response Time = M + (F • V)

This formula gives the time in milliseconds (msec =0.001 sec) required to fill the volume V to 90% of supplypressure or to exhaust the volume V to 10% of supplypressure. M and F are average response constants,and their values for each valve are found in the ROSSmaster catalog. V is the number of cubic inches in thevolume to be filled or exhausted.

What is the constant M? When a valve is energized ittakes a number of milliseconds for the valve to shift andallow a steady flow of air to be established at the outletport. In like manner, when the valve is de-energized ittakes a number of milliseconds for the valve to shift andcut off the flow of outlet air and establish the flow ofexhaust air. These valve “movement” times are desig-nated by M in the above formula.

What is the constant F? After the valve has shifted andair flow to fill the volume V is established, the air flowsrapidly at first, then flows at a reducing rate as pressurebuilds up in the volume V. In exhausting the volume V,air flows rapidly at first, then at a reducing rate as thepressure falls in the volume V. In either case theaverage flow rate is represented in the above formula byF. It is the average number of milliseconds required tofill or exhaust one cubic inch of the volume V. Theproduct F • V, then, is the number of millisecondsrequired to fill or exhaust the entire volume V after thevalve has shifted.

The F values for each valve are found in the ROSSmaster catalog, and they are generally different for thefilling and exhausting functions. The response timetables list the F values under the headings “In-Out” and“Out-Exh.” Clearly the values under the In-Out headingare to be used when a volume is being filled, and thevalues under the Out-Exh heading are used when avolume is being exhausted.

Use of the valve response time formula is illustrated bythe following problems.

Problem A: How long will it take to fill a 100-cubic-inchchamber to 90% of supply pressure using an ISO size2 metal spool valve with single solenoid control?

Solution A: Average response constants for this valveare found in the ROSS master catalog. For the size 2series W60 (metal spool) valve the values are M = 41and F = 1.5. The F value of 2.4 is for the exhaust pathsand so is not needed for this problem.

Putting these M and F values into the valve responsetime formula we have:

Average Response Time = 41 + (1.5)(100)

= 41 + 150

= 191 msec

The volume of 100 cubic inches would be filled to 90%of supply pressure in 191 milliseconds, or less than 2/10of a second.

Problem B: What Series 27, normally closed, 3/2 valveis needed to fill a 1000-cubic-inch volume to 90% ofsupply pressure in less than 1/2 second (500 msec)?

Solution B: Put the given data into the basic formula:

500 = M + (F • 1000)

Now we must find the values of M and F that make theright-hand side of the equation less than 500 msec. Wecan shorten our work by noting that the dominantquantity in the formula is F • 1000. Therefore we mustfind a value of F that will make F • 1000 less than 500,i.e., F must be less than 0.5.

In the ROSS master catalog we find the response timeconstants for Series 27, 3/2, normally closed valves.The first F value that is less than 0.5 is 0.43. The M valueis 11. Putting these values into the basic formula weobtain:

Average Response Time = 11 + (0.43)(1000)

= 11 + 430

= 441 msec

We see that the valve to be selected is the larger of thetwo 1/2-inch valves. In the ROSS master catalog weidentify this valve as model number 2773B4001.

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Determining the C v Required of a Valve Operating aCylinder: The charts on pages 10 and 11 can be usedto determine the valve Cv required to operate cylindersof various sizes. The chart on page 10 shows thedisplaced volume of cylinders with various strokes anddiameters. Knowing the volume of the cylinder and thenumber of strokes per minute the cylinder must make,the chart on page 11 can be used to determine theminimum Cv that the operating valve must have.

Parallel Connection of Pneumatic Components: Ifseveral pneumatic components are connected in paral-lel, the flow coefficient of the combination is simply thesum of the flow coefficients of the several units. Forexample:

Cv1

Cv2

Cv3

Combined Cv = Cv1 + Cv2 + Cv3

Series Connection of Pneumatic Components: Ifseveral pneumatic components are connected in se-ries, the flow coefficient of the combination is a littlemore complicated. For example:

Cv1 Cv2 Cv3

1 1 1 1(Cv)2 (Cv1)2 (Cv2)2 (Cv3)2

Using Cv values must always be tempered with goodjudgement. Flow coefficients are determined in thelaboratory under steady flow conditions, a state thatexists only part of the time in an industrial pneumaticsystem. Some allowance must be made for the valveresponse time required to establish steady flow condi-tions. (See discussion of Valve Response Time on page2.) The use of Cv values is just one area in which theexperience and judgement of the pneumatic designerbecome important.

The flow coefficient (Cv) of a pneumatic device is ameasure of the device's ability to pass air. The amount ofair that can flow through various valves, for example, is indirect proportion to the sizes of their flow coefficients, forgiven inlet and outlet pressures.

Air flow through a valve (or other device) is measuredin the laboratory under carefully controlled conditions.From this measurement plus measurements of the inletand outlet pressures, the flow coefficient can be deter-mined. The relationship between these factors at a tem-perature of 68°F (20°C) is expressed in the followingformula:

Q = 0.98 Cv P(P2 + 14.7)

where Q = air flow in standard cubic feet per minute (scfm)

Cv = flow coefficient

P1 = inlet pressure (psig)

P2 = outlet pressure (psig)

P = pressure drop across valve (psi)= P1 – P2

Knowing the flow coefficient and the inlet and outletpressures, the flow through a valve can be determinedusing this formula.

A simpler way, however, to determine air flow is to use thegraph on the page 5. This graph is based on the aboveformula when Cv = 1. Since air flow is proportional to Cv, this graph can be used for any Cv value; simply multiplythe air flow rate given on the graph by the new Cv value.

Example: Given a supply pressure of 125 psig, a pres-sure drop of 7 psi, and a valve with a Cv of 4.2, find the rateof air flow through the valve.

From the 7 psi pressure drop value at the bottom of thegraph, move upward to the supply pressure line for 125psig. From this point move to the left and read the air flowrate: 30 scfm. Since this is the air flow rate for Cv = 1,multiply by 4.2 to obtain the air flow rate for the valve inthis example. 30 x 4.2 = 126 scfm.

= + +Combined Cv:

Flow Coefficients

ROSS CONTROLS® 8

Standardized Testing

0 = upstream temperature

P1 = pressure at upstream tap

ÐP = pressure drop

D = inside diameter of conductor

The illustration below shows a standardized test set-up established by theAmerican National Standards Institute. ROSS engineering practice calls forstrict adherence to the new testing standards established by ANSI.

The adoption of the new ANSI testing standards by the pneumatics industry willbe a major step forward in eliminating the confusion about Cv ratings which hassometimes existed. Acceptable variations in instrumentation and test hookupcan still yield variations up to 15 per cent in Cv values. However, this will stillrepresent a major improvement over the extreme variations which havesometimes existed in the past.

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Flow Chart for C v = 1

ROSS CONTROLS® 10

Cylinder Volume

7654321

ROSS ControlsTM11

0.1 0.2 0.3 0.4 0.6 0.8 1.0 2.0 3.0 4.0 6.0 8.0 10 20 30 40 50 60 8070

600

400

200

150

80

60

50

40

30

25

20

152

4

6

8

10

100

8

Valve Sizing

ROSS CONTROLS® 12

VALVES

Single Square (Unit forcontrolling flow. Containsinformation on portingand flow paths.

One Flow Path

Two Closed Ports

Two Flow Paths

Two Flow Paths &One Closed Port

Two Flow Paths withCross Connection

One By-pass Flow Path& Two Closed Ports

Two External Ports

2-Position Envelope

3-PositionEnvelope

Two DistinctPositions &One Transi-tory (center)Position

2/2 Valve2 Ports, 2 Positions


FLUID CONDUCTORS

Working Line

Exhaust Line orControl Line

Lines Crossing

Lines Connected

Mechanical Connection(Shaft, Rod, etc.)

Fixed Restriction

Variable Restriction

Flexible Line

Plugged Port

Direction of Flow

ENERGY SOURCES

Pneumatic(Any fluid power source)

Compressor(Fixed displacement)

Vacuum Pump(Fixed displacement)

Accumulator

Pneumatic Symbols

FLUID CONDITIONERS

Basic Envelope(Filters and lubricators)

Filter(Manual drain)

Filter(Automatic drain)

Lubricator(Less drain)

Lubricator(Manual drain)

Air Dryer

Pressure Regulator(Adjustable,self-relieving)

Pressure Regulator(Adjustable,non-relieving)

or

VALVES (cont.)



4/3 Valve4 Ports,3 Positions

5/3 Valve5 Ports,3 Positions

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ENERGYCONVERSION (cont.)

CylinderDouble ActingSingle rod

CylinderDouble ActingDouble rod

CylinderDouble ActingSingle fixed cushion

CylinderDouble ActingTwo adjustablecushions

Quick ActingCoupling

Quick ActingCoupling – Withmechanically openednon-return valve

Quick ActingCoupling – Notcoupled, with open end

Quick ActingCoupling – Notcoupled; closed by freenon-return valve

Silencer

Air MotorOne directional flow

Air MotorTwo directional flows

Release of InternalPilot Pressure

Interior Control Path

Solenoid orInternal Pilot

Solenoid orExternal Pilot

Solenoid plusInternal Pilot

Two External Pilots

Solenoid & Pilot orManual Override

Solenoid & Pilot orManual Overrideand Pilot

DifferentialPressure

ENERGYCONVERSION

CylinderSpring return

CylinderUnspecifiedreturn

ACTUATORSand CONTROLS

Detent – 2 PositionVertical line indicatesflow position.

Muscular Control

Push-button

Lever

Pedal or Treadle

Plunger or PositionIndicator Pin

Spring

Roller

Roller - One way

SolenoidSingle winding

SolenoidTwo opposite windings

Variable SolenoidTwo opposite windings

Pilot PressureExternal

Pilot PressureInternal

Pneumatic Symbols

ROSS CONTROLS® 14

SPECIALPURPOSE VALVES

Check ValveNon-return

Check ValveSpring loaded

Flow Control Valve

Shuttle Valve

SequenceValve

Shutoff Valve

Pilot Control Identification. The following numbers areused on symbols for pilot control or control port identifi-cation. When an electrical or pneumatic signal is appliedto the control or port, the following conditions result:

10 Inlet 1 is closed12 Inlet 1 is connected to outlet 2.14 Inlet 1 is connected to outlet 4.

Use of some of the above controls are shown in thefollowing examples.

Example 1. 2/2 (2 port, 2 position) normally open,spring return valve.

Valve not actuated.

Valve actuated bypneumatic signalapplied to port 10.

MEASURINGINSTRUMENTSand SWITCHES

Pressure Gauge

Pressure Actu-ated ElectricSwitch

Flow Gauge

Pneumatic Symbols — Port Identification

The port markings on ROSS valves are the same asthose shown on the valve symbols in ROSS literature.Main ports are identified by numbers, except for SAEvalves which continue to use letters for identification.

Older valves may not have numerical port markings. Thecorrespondence between the older markings and thecurrent numerical markings is shown below.

Current PortMarking Previous Port Marking

1 IN, P2 OUT 1, OUT A, CYL 1, CYL A3 EXH, EA, R4 OUT 2, OUT B, CYL 2, CYL B5 EXH, EB, S12 CA14 CB

Main Port Identification. Numbers 1 through 5 areused for main port identification as follows:

1 Normal inlet port.2 Normal outlet port.2 & 4 Normal outlet ports where two are

provided.3 Normal exhaust port.3 & 5 Normal exhaust ports where two are

provided.

PORT IDENTIFICATION

10

2

1

10

2

1

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Example 2. 3/2 (3 port, 2 position) normally closed,spring return valve.

Valve not actuated.

Valve actuated byelectrical signalapplied to control 12.

Example 3. 4/2 (4 port, 2 position) spring return valve.

Valve not actuated.

Valve actuated bypneumatic signalapplied to control 14.

Port Identification — Conversion Factors

Example 4. 5/3 (5 port, 3 position) closed center, spring centered valve.

Valve in non-actuatedcenter position.



Auxiliary Ports for Pilot Operators

Port Function

X External pilot supply portX1Y2Y3 Exhaust port for solenoid pilotY4

Example:

24

1 35

14 12

24

1 35

14 12

24

1 35

14 12

13

2

12

13

2

12

13

24

14

13

24

14

CONVERSION FACTORS

U.S. to Metric ConversionFrom To Multiply By

inches .................. millimeters .................... 25.40

feet ...................... meters ............................ 0.3048

miles .................... kilometers ....................... 1.609

square feet .......... square meters ................ 0.09290

cubic feet ............. cubic meters ................... 0.02832

cubic feet ............. liters .............................. 28.32

gallon ................... liters ................................ 3.785

pounds ................ kilograms ........................ 0.4536

lb/in2 (psi) ............. bar .................................. 0.06895

ft3/min .................. liters/second (l/s) ............ 0.4718

Metric to U.S. ConversionFrom To Multiply By

millimeters ........... inches ............................. 0.03937

meters ................. feet ................................. 3.281

kilometers ............ miles ............................... 0.6214

square meters ..... square feet ................... 10.76

cubic meters ........ cubic feet ...................... 35.30

liters ..................... cubic feet ........................ 0.03531

liters ..................... gallons ............................ 0.2642

kilograms ............. pounds ........................... 2.205

bar ....................... lb/in2.................................................... 14.50

liters/second (l/s) . ft3/min ............................. 2.120

13

24

14

XY4

}}

Form A10072Rev. 6/96 © Copyright 1996, ROSS CONTROLS.

Warranty

ROSS CONTROLS®

P.O. Box 7015Troy, Michigan 48007 U.S.A.Telephone (00) 1-248-764-1800Fax (00) 1-248-764-1850www.rosscontrols.comIn the United States:Customer Service: 1-800-GET-ROSSTechnical Service: 1-800-TEK-ROSSROSS/FLEX® Service: 1-888-ROSS-FLX

Products manufactured by ROSS are warranted to be free of defects in material and workmanship for a period of one year from thedate of purchase. ROSS’ obligation under this warranty is limited to repair or replacement of the product or refund of the purchaseprice paid solely at the discretion of ROSS and provided such product is returned to ROSS freight prepaid and upon examination byROSS is found to be defective. This warranty shall be void in the event that product has been subject to misuse, misapplication, impropermaintenance, modification or tampering. THE WARRANTY EXPRESSED ABOVE IS IN LIEU OF AND EXCLUSIVE OF ALL OTHERWARRANTIES AND ROSS EXPRESSLY DISCLAIMS ALL OTHER WARRANTIES EITHER EXPRESSED OR IMPLIED WITHRESPECT TO MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. ROSS MAKES NO WARRANTY WITH RESPECTTO ITS PRODUCTS MEETING THE PROVISIONS OF ANY GOVERNMENTAL OCCUPATIONAL SAFETY AND/OR HEALTH LAWSOR REGULATIONS. IN NO EVENT SHALL ROSS BE LIABLE TO PURCHASER, USER, THEIR EMPLOYEES OR OTHERS FORINCIDENTAL OR CONSEQUENTIAL DAMAGES WHICH MAY RESULT FROM A BREACH OF THE WARRANTY DESCRIBEDABOVE OR THE USE OR MISUSE OF THE PRODUCTS. NO STATEMENT OF ANY REPRESENTATIVE OR EMPLOYEE OF ROSSSHALL EXTEND THE LIABILITY OF ROSS AS SET FORTH HEREIN.

ROSS EUROPA GmbHRobert-Bosch-Stra βββββe 2D-63225 Langen, GermanyTelephone (011) 49-6103-7597-0Fax (011) 49-6103-7469-4

ROSS ASIA K.K.10209-5 Tana, Sagamihara-shiKanagawa 229-1124, JapanTelephone (011) 81-427-78-7251Fax (011) 81-427-78-7256

ROSS UK Ltd.St. James Road, BrackleyNorthamptonshire NN13 7XYUnited KingdomTelephone (011) 44-1280-706668Fax (011) 44-1280-705630

ROSS ASIA K.K.-CHINA LIAISON OFFICERoom 17B, FuHai Building,288 Huanghe RoadShanghai, ChinaTelephone (011) 21-6372-2579Fax (011) 21-6372-2505ROSS CONTROLS INDIA Pvt. Ltd.

‘B’ Mount Chambers, Fourth Floor758 Mount RoadChennai, 600 002 IndiaTelephone (011) 91-44-841-3136Fax (011) 91-44-841-3137

ROSS SOUTH AMERICA Ltda.Rua Olavo Gonçalves, 43Centro- São Bernardo Do CampoSão Paulo, Brazil CEP 09725-020Telephone (011) 55-11-9122-2705

Documents

Pneumatica ROSS